Download Section 8.4 – Trigonometric Identities

107
Section 8.4 – Trigonometric Identities
Definition
Two functions f and g are said to be identically equal if f(x) = g(x) for all x
in the domain of f and g. We say that f(x) = g(x) is an identity.
Determine if the following are identities.
Ex. 1a
(3x + 2)2 = 9x2 +12x + 4
Ex. 1b
Ex. 1c
4x – 5 = 23
Ex. 1d
tan2(θ) + 1 = sec2(θ)
(x +3)(x −2)
(x−2)
=x+3
Solution:
a)
Yes, since if we expand (3x + 2)2, we get 9x2 +12x + 4. Both
expressions are defined for all real numbers.
b)
Yes, we established that identity in the previous chapter.
c)
No, since 4x – 5 only equals 23 when x = 7. It is not true for all
real numbers.
d)
These two functions are identical except at x = 2.
(undefined ≠ 5). We will see in Calculus, we can fix this
problem, be defining a piecewise defined function.
We have already established quite a few trigonometric identities that you
should already know.
Fundamental Identities
tan(θ) =
csc(θ) =
1
sin(θ)
sin(θ)
cos(θ)
sec(θ) =
cos(θ)
sin(θ)
cot(θ) =
1
cos(θ)
sin2(θ) + cos2(θ) €
=1
tan2(θ) + 1 = sec2€(θ)
sin(– θ) = – sin(θ)
€
csc(– θ) = – csc(θ)
cos(– θ) = cos(θ)
€
sec(– θ) = sec(θ)
cot(θ) =
1
tan(θ)
cot2(θ) + 1 = csc2(θ)
tan(– θ) = – tan(θ)
€
cot(– θ) = – cot(θ)
Keep in mind that we may be working with a variation of one of these
identities. For instance, instead of using sec(θ) =
use cos(θ) =
1
sec(θ)
1
,
cos(θ)
we might need to
or instead of cot2(θ) + 1 = csc2(θ), we might need to use
cot2(θ) – csc2(θ) = – 1. To prove an identity, we will need to establish a list
€
of tools we can employ to simplify trigonometric
expressions.
€
108
Objective 1:
Use Algebra to Simplify Trigonometric Expressions.
There are six major tools we use when working trigonometric expressions.
Below is a list each tool and an example to see how the tool is used.
Tools
1)
Rewriting all the trigonometric functions in terms of sine and cosine.
cos(θ)
1
Example: cot(θ)•sec(θ)
(cot(θ) =
& sec(θ) =
)
sin(θ)
=
=
2)
cos(θ)
1
•
sin(θ) cos(θ)
1
= csc(θ)
sin(θ)
€
€
Multiplying by a conjugate to make a difference of squares and using
€
€ identity:
a Pythagorean
sin(θ)
Example:
(the conjugate of 1 + cos(θ) is 1 – cos(θ))
€
1+cos(θ)
sin(θ)
1−cos(θ)
=
•
1+cos(θ) 1−cos(θ)
sin(θ)−sin(θ)cos(θ)
=
€
€
€
€
3)
cos(θ)
=
(sin2(θ) = 1 – cos2(θ))
2
1−cos (θ)
sin(θ)−sin(θ)cos(θ)
=€
=
(expand)
2
sin (θ)
sin(θ)
sin(θ)cos(θ)
–
sin2 (θ)
sin2 (θ)
1
cos(θ)
–
sin(θ)
sin(θ)
(divide both terms by sin2(θ))
(reduce)
(csc(θ) =
1
sin(θ)
& cot(θ) =
cos(θ)
sin(θ)
= csc(θ) – cot(θ)
€
€
Finding a common
denominator and adding the fractions together.
csc(θ)
csc(θ)
Example:
+
€
€
€
€
csc(θ)−cot(θ)
csc(θ)+cot(θ)
LCD = (csc(θ) – cot(θ))(csc(θ) + cot(θ))
=
€
=
€
€
€
€
=
=
csc(θ)
csc(θ)+cot(θ)
•
csc(θ)−cot(θ) csc(θ)+cot(θ)
2
csc€
(θ)+csc(θ)cot(θ)
csc2 (θ)−cot2 (θ)
2csc
€
2
2
(θ)
2
csc (θ)−cot (θ)
2csc2 (θ)
1
+
+
csc(θ)
csc(θ)−cot(θ)
•
csc(θ)+cot(θ) csc(θ)−cot(θ)
csc2 (θ)−csc(θ)cot(θ)
csc2 (θ)−cot2 (θ)
€ (1 = csc2€
(θ) – cot2(θ))
€
= 2csc2(θ)
)
109
4)
Expanding and simplifying.
Example:
€
5)
=
(tan(θ)+1)(tan(θ)+1)−sec2 (θ)
tan(θ)
=
tan2 (θ)+2 tan(θ)+1−sec2 (θ)
tan(θ)
€
=
€
=
tan2 (θ)−sec2 (θ)+2 tan(θ)+1
tan(θ)
−1+2 tan(θ)+1
tan(θ)
2 tan(θ)
=2
tan(θ)
(tan2(θ) – sec2(θ) = – 1)
=
€
Factoring and simplifying.
€
cos2 (θ)−1
Example:
(factor the numerator and denominator)
cos2 (θ)−cos(θ)
€
(cos(θ)+1)(cos(θ)−1)
=
(reduce)
=
€
€
6)
(tan(θ)+1)2 −sec2 (θ)
tan(θ)
=
cos(θ)(cos(θ)−1)
(cos(θ)+1)
cos(θ)
cos(θ)
1
+
cos(θ)
cos(θ)
(divide by cos(θ))
(sec(θ) =
1
)
cos(θ)
= 1 + sec(θ)
€
Multiplying
top and bottom of a fraction by a form of one:
€
Example:
1+
€tan(θ)
1− tan(θ)
=
€
1
tan(θ)
1
tan(θ)
•
+
−
€cot(θ) + 1
1
tan(θ)
1
tan(θ)
tan(θ)
tan(θ)
tan(θ)
tan(θ)
€
(
tan(θ)
tan(θ)
= 1 and cot(θ) =
1
)
tan(θ )
cot(θ) − 1
€
Objective 2:
€
Establishing Identities
€
€ an identity, we begin with one side of the identity, usually the
To establish
more complicated side, and manipulate that side until we get the same
expression as we have on the other side. It is important in this process is to
keep examining your goal and try to use our tools to get closer to the goal.
110
Establish the identity:
Ex. 2
sin(θ)csc(θ) – cos2(θ) = sin2(θ)
Solution:
The left side is more complicated. Since we have mostly sine and
cosine functions, lets use our first tool to write everything in terms of
sine and cosine.
sin(θ)csc(θ) – cos2(θ)
= sin(θ)•
1
sin(θ)
– cos2(θ)
= 1 – cos2(θ)
= sin2(θ)
Ex. 3
(csc(θ) =
1
)
sin(θ)
(simplify)
(sin2(θ) + cos2(θ) = 1 ⇒ 1 – cos2(θ) = sin2(θ))
€ established.
The identity has been
€
1 = (1 – cos2(– θ))(1 + cot2(θ))
Solution:
The right side is more complicated, so we will start with that side.
(1 – cos2(– θ))(1 + cot2(θ)) (cos(– θ) = cos(θ), so cos2(– θ) = cos2(θ))
= (1 – cos2(θ))(1 + cot2(θ)) (expand)
= 1 + cot2(θ) – cos2(θ) – cos2(θ)cot2(θ) (use sine & cosine)
=1+
cos2 (θ)
sin2 (θ)
– cos2(θ) – cos2(θ)
cos2 (θ)
sin2 (θ)
We seem to have reached a dead end. Let's go back to before we
expanded and try a different approach:
= (1 – cos2(θ))(1 + cot2(θ))
€
But, 1 – cos2(θ)) =€sin2(θ) and 1 + cot2(θ) = csc2(θ)
= (sin2(θ))(csc2(θ))
= (sin2(θ))(
=1
Ex. 4 €
1
2
sin (θ)
)
(csc(θ) =
1
)
sin(θ)
(reduce)
The Identity has been established.
€
1–
sin2 (−θ)
1−cos(−θ)
= – cos(θ)
Solution:
We will begin with the left side.
1€–
sin2 (−θ)
1−cos(−θ)
(sin(– θ) = – sin(θ), but sin2(– θ) = sin2(θ) & cos(– θ) = cos(θ))
€
111
=1–
=
€ =
sin2 (θ)
1−cos(θ)
1−cos(θ)
1−cos(θ)
–
(get a common denominator and subtract)
sin2 (θ)
1−cos(θ)
1−cos(θ)−sin2 (θ)
1−cos(θ)
(regroup the numerator)
1−sin2 (θ)−cos(θ)
1−cos(θ)
(1 – sin2(θ) = cos2(θ))
€
= €
€
=
€
=
cos2 (θ)−cos(θ)
1−cos(θ)
−cos(θ)(1−cos(θ))
1−cos(θ)
= – cos(θ)
(factor out a – cos(θ) in the numerator)
(reduce)
The identity has been established.
€
cos(x)
1+sin(x)
€Ex. 5
+
1+sin(x)
cos(x)
= 2sec(x)
Solution:
cos(x)
+
€
1+sin(x)
(get a common
1+sin(x)
cos(x)
cos(x)
1+sin(x) 1+sin(x)
€ cos(x)
=
•
+
•
1+sin(x) cos(x)
cos(x)
1+sin(x)
cos2 (x)
=€
+
cos(x)(1+sin(x))
€
denominator and add)
(expand the numerator)
1+2sin(x)+sin2 (x)
cos(x)(1+sin(x))
1
€
€
€
€
€
€
If€we examine
€ the right
€ side, we see the sec(x) = cos(x) so we do not
want to expand the denominator, but try to get rid of 1 + sin(x).
2
€
cos2 (x)+1+2sin(x)+sin
(x)
=
(regroup the numerator)
cos(x)(1+sin(x))
€
cos2 (x)+sin2 (x)+1+2sin(x)
=
(sin2(x) + cos2(x) = 1)
=
=
=
=
cos(x)(1+sin(x))
1+1+2sin(x)
cos(x)(1+sin(x))
2+2sin(x)
cos(x)(1+sin(x))
2(1+sin(x))
cos(x)(1+sin(x))
2
cos(x)
= 2sec(x)
(factor 2 out of the numerator)
(reduce)
(sec(x) =
The identity has been established.
€
€
1
)
cos(x)
€
112
sin(w)cos(w)
Ex. 6
2
=
2
cos (w)−sin (w)
tan(w)
1− tan2 (w)
Solution:
First, rewrite the right side in terms of sine and cosine:
tan(w)
€
2
1− tan (w)
€
1−
€
=
=
€
=
€
Ex. 7
€
€
(multiply top and bottom by cos2(w))
1−
cos2 ( w )
cos2 ( w )
•
cos2 ( w )
1
2
cos ( w )
1
sin( w ) cos2 ( w )
•
cos( w )
1
2
2
cos ( w )
sin ( w ) cos2 ( w )
€1
€
=
sin( w )
cos( w )
sin2 ( w )
=
sin( w )
cos(
€ w)
sin2 ( w )
−
•
cos2 ( w )
sin( w ) cos( w )
•
1
1
2
2
cos ( w )
sin ( w ) 1
−
•
1
1
1
1
sin(w)cos(w)
The identity has been established.
cos2 (w) − sin2 (w)
sin3 (t)+cos3 (t)
sec(t)−sin(t)
tan(t)−1
=
2
1−2cos (t)
Solution:
We need to reduce the powers of the sine and cosine function on the
left side. If we
€ examine this carefully, it is a sum of cubes.
sin3 (t)+cos3 (t)
(F3 + L3 = (F + L)(F2 – FL + L2))
2
1−2cos (t)
=
€
=
€
=
€
=
€
=
€
€
(reduce)
(sin(t)+cos(t))(sin2 (t)−sin(t)cos(t)+cos2 (t))
1−2cos2 (t)
(sin(t)+cos(t))(sin2 (t)+cos2 (t)−sin(t)cos(t))
2
1−2cos (t)
(sin(t)+cos(t))(1−sin(t)cos(t))
1−2cos2 (t)
(sin(t)+cos(t))(1−sin(t)cos(t))
2
2
1−cos (t)−cos (t)
(sin(t)+cos(t))(1−sin(t)cos(t))
sin2 (t)−cos2 (t)
(regroup the numerator)
(sin2(t) + cos2(t) = 1)
(now, let's work with the denominator)
(sin2(t) = 1 – cos2(t))
(factor the denominator)
113
=
=
€
=
(1−sin(t)cos(t))
(sin(t)−cos(t))
=
sin( t) cos( t)
1
−
cos( t)
cos( t)
sin( t)
cos( t)
−
cos( t)
cos( t)
=
sec(t) − sin(t)
tan(t) − 1
€
=
€
=
€
=
€
=
€
=
=
=
€
=
€
€
€
(tan(t) =
sin(t)
cos(t)
and sec(t) =
1
)
cos(t)
The identity has been established.
€
= sec(θ) + tan(θ)
€
[1 +cos(θ)]+sin(θ)
[1 +cos(θ)]−sin(θ)
[1 +cos(θ)]+sin(θ)
[1 +cos(θ)]−sin(θ)
(multiply top & bottom by [1 + cos(θ)] + sin(θ))
•
[1 +cos(θ)]+sin(θ)
[1 +cos(θ)]+sin(θ)
[1 +cos(θ)]2 +2[1 +cos(θ)]sin(θ)+sin2 (θ)
[1 +cos(θ)]2 −sin2 (θ)
(expand)
(expand)
2
1 +2cos(θ)+cos
(θ)+2sin(θ)+2cos(θ)sin(θ)+sin2 (θ)
€
1 +2cos(θ)+cos2 (θ)−sin2 (θ)
(regroup)
sin2 (θ)+cos2 (θ)+1 +2cos(θ)+2sin(θ)+2cos(θ)sin(θ)
1 −sin2 (θ)+2cos(θ)+cos2 (θ)
(sin2(θ) + cos(θ) = 1 and 1 – sin2(θ) = cos2(θ))
€
€
•
1
)
cos(t)
Solution:
Looking at the left side, we need to try to get rid of the denominator.
1 +cos(θ)+sin(θ)
(group 1 + cos(θ))
€ 1 +cos(θ)−sin(θ)
=
€
at the right side, we need tan(t) – 1 in the
1
cos( t)
1
cos( t)
1 +cos(θ)+sin(θ)
1 +cos(θ)−sin(θ)
€Ex. 8
€
(reduce)
denominator. Thus, we need to multiply top and bottom by
€
€
(sin(t)+cos(t))(1−sin(t)cos(t))
(sin(t)+cos(t))(sin(t)−cos(t))
(1−sin(t)cos(t))
(looking
(sin(t)−cos(t))
1+1 +2cos(θ)+2sin(θ)+2cos(θ)sin(θ)
cos2 (θ)+2cos(θ)+cos2 (θ)
2+2cos(θ)+2sin(θ)+2cos(θ)sin(θ)
2cos(θ)+2cos2 (θ)
2[1+cos(θ)+sin(θ)+cos(θ)sin(θ)]
2[cos(θ)+cos2 (θ)]
1+cos(θ)+sin(θ)+cos(θ)sin(θ)
cos(θ)+cos2 (θ)
(combine like terms)
(factor out 2 from the top & bottom)
(reduce)
(factor the top by grouping)
114
[1+cos(θ)]+sin(θ)[1+cos(θ)]
=
=
€
€
€
=
=
=
cos(θ)+cos2 (θ)
[1+cos(θ)][1+sin(θ)]
cos(θ)+cos2 (θ)
[1+cos(θ)][1+sin(θ)]
cos(θ)[1+cos(θ)]
[1+sin(θ)]
cos(θ)
1
sin(θ)
+
cos(θ)
cos(θ)
(factor out cos(θ) in the denominator)
(reduce)
(divide)
(sec(θ) =
1
,
cos(θ)
tan(θ) =
= sec(θ) + tan(θ)
€
€
€
€
€
sin(θ)
)
cos(θ)